| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Angles between vectors |
| Difficulty | Standard +0.3 This is a straightforward vectors question requiring standard techniques: calculating magnitudes, using the scalar product formula for angles, finding a midpoint, and solving a distance equation. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Show that \(OA = OB = \sqrt{5}\) | B1 | CWO |
| Evaluate the scalar product of the correct position vectors | M1 | e.g. \((0-1+0)\); condone use of \(AO\) and/or \(BO\) |
| Divide *their* scalar product by the product of the moduli of *their* vectors and evaluate the inverse cosine of the result | M1 | Must reach an angle. The question asks for use of scalar product, so alternative methods (e.g. cosine rule) are not accepted. |
| Obtain answer \(101.5°\) | A1 | Accept \(102°\) or better. Mark radians \((1.77)\) as a misread. Do not ISW: \(78.5°\) as final answer scores A0. |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(M\) has position vector \(\mathbf{i} - \mathbf{k}\) | B1 | OE |
| Taking a general point of \(OM\) to have position vector \(\lambda\mathbf{i} - \lambda\mathbf{k}\), express \(AP = \sqrt{7}\, OA\) as an equation in \(\lambda\) | *M1 | \(\lambda(\text{their } \overrightarrow{OM})\) |
| State a correct equation in any form | A1 | e.g. \(\sqrt{(-2+\lambda)^2 + 1 + (-\lambda)^2} = \sqrt{7}\sqrt{5}\) |
| Reduce to \(\lambda^2 - 2\lambda - 15 = 0\) | A1 | OE |
| Solve a quadratic and state a position vector | DM1 | |
| Obtain answers \(5\mathbf{i} - 5\mathbf{k}\) and \(-3\mathbf{i} + 3\mathbf{k}\) | A1 | Accept coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(OP = \gamma\sqrt{2}\) | B1 | |
| State or imply \(\cos\frac{1}{2}AOB = \sqrt{\frac{2}{5}}\) and use cosine rule to form equation in \(\gamma\) | *M1 | Allow \(\cos\frac{1}{2}AOB = 0.632\ldots\) |
| State a correct equation in any form | A1 | e.g. \(35 = 5 + 2\gamma^2 - 2\sqrt{5}\cdot\gamma\sqrt{2}\cdot\frac{\sqrt{2}}{\sqrt{5}}\) |
| Reduce to \(\gamma^2 - 2\gamma - 15 = 0\) | A1 | OE |
| Solve a quadratic and state a position vector | DM1 | |
| Obtain answers \(5\mathbf{i} - 5\mathbf{k}\) and \(-3\mathbf{i} + 3\mathbf{k}\) | A1 | Accept coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(M\) has position vector \(\mathbf{i} - \mathbf{k}\) | B1 | OE |
| State or imply \(AM = \sqrt{3}\) | B1 | |
| Use Pythagoras to find \(MP\) | *M1 | \(MP = \sqrt{35 - (AM)^2}\) |
| Obtain \(MP = 4\sqrt{2}\) | A1 | |
| Correct method to find a position vector | DM1 | \((\mathbf{i}-\mathbf{k}) \pm 4(\mathbf{i}-\mathbf{k})\) |
| Obtain answers \(5\mathbf{i} - 5\mathbf{k}\) and \(-3\mathbf{i} + 3\mathbf{k}\) | A1 | Accept coordinates |
| Total | 6 |
## Question 11(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Show that $OA = OB = \sqrt{5}$ | B1 | CWO |
| Evaluate the scalar product of the correct position vectors | M1 | e.g. $(0-1+0)$; condone use of $AO$ and/or $BO$ |
| Divide *their* scalar product by the product of the moduli of *their* vectors and evaluate the inverse cosine of the result | M1 | Must reach an angle. The question asks for use of scalar product, so alternative methods (e.g. cosine rule) are not accepted. |
| Obtain answer $101.5°$ | A1 | Accept $102°$ or better. Mark radians $(1.77)$ as a misread. Do not ISW: $78.5°$ as final answer scores A0. |
| **Total** | **4** | |
---
## Question 11(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $M$ has position vector $\mathbf{i} - \mathbf{k}$ | B1 | OE |
| Taking a general point of $OM$ to have position vector $\lambda\mathbf{i} - \lambda\mathbf{k}$, express $AP = \sqrt{7}\, OA$ as an equation in $\lambda$ | *M1 | $\lambda(\text{their } \overrightarrow{OM})$ |
| State a correct equation in any form | A1 | e.g. $\sqrt{(-2+\lambda)^2 + 1 + (-\lambda)^2} = \sqrt{7}\sqrt{5}$ |
| Reduce to $\lambda^2 - 2\lambda - 15 = 0$ | A1 | OE |
| Solve a quadratic and state a position vector | DM1 | |
| Obtain answers $5\mathbf{i} - 5\mathbf{k}$ and $-3\mathbf{i} + 3\mathbf{k}$ | A1 | Accept coordinates |
**Alternative method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $OP = \gamma\sqrt{2}$ | B1 | |
| State or imply $\cos\frac{1}{2}AOB = \sqrt{\frac{2}{5}}$ and use cosine rule to form equation in $\gamma$ | *M1 | Allow $\cos\frac{1}{2}AOB = 0.632\ldots$ |
| State a correct equation in any form | A1 | e.g. $35 = 5 + 2\gamma^2 - 2\sqrt{5}\cdot\gamma\sqrt{2}\cdot\frac{\sqrt{2}}{\sqrt{5}}$ |
| Reduce to $\gamma^2 - 2\gamma - 15 = 0$ | A1 | OE |
| Solve a quadratic and state a position vector | DM1 | |
| Obtain answers $5\mathbf{i} - 5\mathbf{k}$ and $-3\mathbf{i} + 3\mathbf{k}$ | A1 | Accept coordinates |
**Alternative method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $M$ has position vector $\mathbf{i} - \mathbf{k}$ | B1 | OE |
| State or imply $AM = \sqrt{3}$ | B1 | |
| Use Pythagoras to find $MP$ | *M1 | $MP = \sqrt{35 - (AM)^2}$ |
| Obtain $MP = 4\sqrt{2}$ | A1 | |
| Correct method to find a position vector | DM1 | $(\mathbf{i}-\mathbf{k}) \pm 4(\mathbf{i}-\mathbf{k})$ |
| Obtain answers $5\mathbf{i} - 5\mathbf{k}$ and $-3\mathbf{i} + 3\mathbf{k}$ | A1 | Accept coordinates |
| **Total** | **6** | |11 With respect to the origin $O$, the points $A$ and $B$ have position vectors given by $\overrightarrow { O A } = 2 \mathbf { i } - \mathbf { j }$ and $\overrightarrow { O B } = \mathbf { j } - 2 \mathbf { k }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $O A = O B$ and use a scalar product to calculate angle $A O B$ in degrees.\\
The midpoint of $A B$ is $M$. The point $P$ on the line through $O$ and $M$ is such that $P A : O A = \sqrt { 7 } : 1$.
\item Find the possible position vectors of $P$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q11 [10]}}